Question:

Step 1: Identify the derivative of the denominator. We compute the derivative of the denominator $x^4 + x + 1$: $$\frac{d}{dx}(x^4 + x + 1) = 4x^3 + 1$$ This will help us recognize a pattern in the integrand. Step 2: Find a function whose derivative matches the integrand. We attempt to find a function of the form $\frac{x}{x^4 + x + 1}$ and compute its derivative using the quotient rule: $$\frac{d}{dx}\left(\frac{x}{x^4+x+1}\right) = \frac{(x^4+x+1) \cdot 1 - x \cdot (4x^3+1)}{(x^4+x+1)^2}$$ Step 3: Simplify the derivative. Expanding the numerator: $$= \frac{x^4 + x + 1 - 4x^4 - x}{(x^4+x+1)^2} = \frac{-3x^4 + 1}{(x^4+x+1)^2}$$ Step 4: Relate this to the original integrand. Notice that: $$\frac{3x^4 - 1}{(x^4+x+1)^2} = -\frac{d}{dx}\left(\frac{x}{x^4+x+1}\right)$$ This means our integrand is the negative of the derivative we just found. Step 5: Integrate to find $f(x)$. Since the integrand is the negative of the derivative, we have: $$f(x) = \int \frac{3x^4 - 1}{(x^4+x+1)^2}\, dx = -\frac{x}{x^4+x+1} + C$$ where $C$ is the constant of integration. Step 6: Use the initial condition to find $C$. Given that $f(0) = 0$: $$f(0) = -\frac{0}{0^4+0+1} + C = 0 + C = 0$$ Therefore, $C = 0$. Step 7: Evaluate $f(-1)$. With $C = 0$, we have $f(x) = -\frac{x}{x^4+x+1}$. Substituting $x = -1$: $$f(-1) = -\frac{-1}{(-1)^4 + (-1) + 1} = \frac{1}{1 - 1 + 1} = \frac{1}{1} = 1$$ **Final Answer:** $f(-1) = 1$, which corresponds to **Option 3**.